正则表达式获取不带参数的 url 的最后一部分(如果存在) [英] Regexp get last part of url without parameters if they exists

问题描述

我正在寻找要在我的 javascript 代码中使用的正则表达式，它为我提供不带参数的 url 的最后一部分 如果它们存在 - 这是示例 - 带和不带参数:

https://contentsfra>-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg

在这两种情况下，我都想得到:

14238253_132683573850463_7287992614234853254_n.jpg

解决方案

这是正则表达式

.*\/([^?]+)

和JS代码:

let lastUrlPart =/.*\/([^?]+)/.exec(url)[1];

let lastUrlPart = url =>/.*\/([^?]+)/.exec(url)[1];//测试让 t1 = "https://scontent-fra3-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg?oh=fdbf6800f853bec5e78a6835c8c8538c87a8c538c8738c8538c873853573850463_72879926142387让 t2 = "https://scontent-fra3-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg"console.log(lastUrlPart(t1));console.log(lastUrlPart(t2));

可能有更好的选择吗?

I looking for regular expression to use in my javascript code, which give me last part of url without parameters if they exists - here is example - with and without parameters:

https://scontent-fra3-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg?oh=fdbf6800f33876a86ed17835cfce8e3b&oe=599548AC

https://scontent-fra3-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg

In both cases as result I want to get:

14238253_132683573850463_7287992614234853254_n.jpg

解决方案

Here is this regexp

.*\/([^?]+)

and JS code:

let lastUrlPart = /.*\/([^?]+)/.exec(url)[1];

let lastUrlPart = url => /.*\/([^?]+)/.exec(url)[1];



// TEST

let t1 = "https://scontent-fra3-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg?oh=fdbf6800f33876a86ed17835cfce8e3b&oe=599548AC"

let t2 = "https://scontent-fra3-1.xx.fbcdn.net/v/t1.0-9/14238253_132683573850463_7287992614234853254_n.jpg"

console.log(lastUrlPart(t1));
console.log(lastUrlPart(t2));

May be there are better alternatives?

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